a) \(x^2-5xy+6y^2\)

\(=x^2-3xy-2xy+6y^2\)

\(=x\left(x-3y\right)-2y\left(x-3y\right)\)

\(=\left(x-2y\right)\left(x-3y\right)\)

b) \(16\left(x-1\right)^2-36y^2\)

\(=\left(4x-4\right)^2-\left(6y\right)^2\)

\(=\left(4x+6y-4\right)\left(4x-6y-4\right)\)

c) \(4\left(x+y\right)-12\left(x+y\right)^2\)

\(=\left(x+y\right)\left[4-12\left(x+y\right)\right]\)

\(=4\left(x+y\right)\left[1-3x-3y\right]\)

\(a.=4a\left(a-2\right)\)

\(b.=x\left(x+2xy+y^2-z^2\right)\)

\(=x\left(\left(x+y\right)^2-z^2\right)\)

\(=x\left(x+y-z\right)\left(x+y+z\right)\)

Gọi f_{( x )} = x³ - x² - 11x + m

      g_{( x )}  = x - 3

Cho g_{( x )} = 0

\(\Rightarrow\)x - 3 = 0

\(\Rightarrow\)x      = 3

\(\Rightarrow\)f_{( 3 )} = 3³ - 3² - 11.3 + m

\(\Rightarrow\)f_{( 3 )} = - 15 + m

Để f_{( x )} \(⋮\)g_{( x )}

\(\Leftrightarrow\)- 15 + m = 0

\(\Rightarrow\)m              = - 15 

Vậy : m = - 15 thì M = x³ - x² - 11x + m \(⋮\)x - 3

Ta có : ( a - b )²  + 4ab

= a² - 2ab + b² + 4ab

= a² + 2ab + b²

= ( a + b )² ( Vế trái )

Do đó : ( a + b )² = ( a - b )² + 4ab 

+) Biến đổi vế phải ta có :

\(\left(A-B\right)^2+4AB\)

\(=A^2-2AB+B^2+4AB\)

\(=A^2+2AB+B^2=\left(A+B\right)^2=VT\left(đpcm\right)\)

(x + 3)^2 = 9(2x - 1)^2

=> x^2 + 6x + 9 = 9(4x^2 - 4x + 1) 

=> x^2 + 6x + 9 = 36x^2 - 36x + 9

=> x^2 - 36x^2 + 6x + 36x + 9 - 9 = 0

=> -35x^2 + 42x = 0

=> x(-35x + 42) = 0

=> x = 0 hoặc -35x + 42 = 0

=> x = 0 hoặc x = 42/35

\(\left(x+3\right)^2=9\left(2x-1\right)^2\)

\(\left(x+3\right)^2=3^2\cdot\left(2x-1\right)^2\)

\(\left(x+3\right)^2=\left(6x-3\right)^2\)

\(\left(x+3\right)^2-\left(6x-3\right)^2=0\)

\(\left(x+3-6x+3\right)\cdot\left(x+3+6x-3\right)=0\)

\(\left(6-5x\right)\cdot7x=0\)

\(\orbr{\begin{cases}6-5x=0\\7x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=0\end{cases}}}\)

\(a^3+b^3+2019ab\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+2019ab\)

\(=673^3-2019ab+2019ab\)

\(=673^3=47832147\)